1988 AMC 12 Problem 18

Attempt Problem 18 of the 1988 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1988 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

18.

At the end of a professional bowling tournament, the top 55 bowlers have a play-off. First #55 bowls #4.4. The loser receives 55th prize and the winner bowls #33 in another game. The loser of this game receives 44th prize and the winner bowls #2.2. The loser of this game receives 33rd prize and the winner bowls #1.1. The winner of this game gets 11st prize and the loser gets 22nd prize. In how many orders can bowlers #11 through #55 receive the prizes?
1010
1616
2424
120120
none of these
Answer: B
Concepts:countingbinary choicestournament outcomes
Small hint:
Exactly four games are played
Big hint:
Each sequence of winners determines a different prize order
Solution:
Each of the four games has two possible winners. A sequence of four outcomes uniquely determines the five prize positions, so there are 24=162^4=16 orders.
Thus the correct answer is B.
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